Problem: Find $\lim_{h\to 0}\dfrac{\dfrac{5}{(2+h)^2}-\dfrac{5}{(2)^2}}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac54$ (Choice B) B $\dfrac54$ (Choice C) C $5$ (Choice D) D The limit doesn't exist
Answer: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $\dfrac{5}{(2+h)^2}-\dfrac{5}{(2)^2}$, we can tell that the function is $f(x)=\dfrac{5}{x^2}$ and the $x$ -value is $2$. In other words, the limit expression is equal to $f'(2)$ for $f(x)=\dfrac{5}{x^2}$. Let's find $f'(x)$ : $f'(x)=5\cdot -2x^{-3}=-\dfrac{10}{x^3}$ Now let's evaluate $f'(2)$ : $f'(2)=-\dfrac{10}{(2)^3}=-\dfrac54$ In conclusion, $\lim_{h\to 0}\dfrac{\dfrac{5}{(2+h)^2}-\dfrac{5}{(2)^2}}{h}=-\dfrac54$.